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If $e$ and $e^{\prime}$ are eccentricities of hyperbola and its conjugate respectively, then
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$\left(\frac{1}{e}\right)^2+\left(\frac{1}{e^{\prime}}\right)^2=1$
Let hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ....(i)
Then its conjugate will be, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$ ....(ii)
If $e$ is eccentricity of hyperbola (i), then $b^2=a^2\left(e^2-1\right)$
or $\frac{1}{e^2}=\frac{a^2}{\left(a^2+b^2\right)}$ ....(iii)
Similarly if $e^{\prime}$ is eccentricity of conjugate (ii), then $a^2=b^2\left(e^{\prime 2}-1\right)$ or $\frac{1}{e^{\prime 2}}=\frac{b^2}{\left(a^2+b^2\right)}$ .....(iv)
Adding (iii) and (iv),
\(\left(\frac{1}{e}\right)^2+\left(\frac{1}{e^1}\right)^2=1\)
Then its conjugate will be, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$ ....(ii)
If $e$ is eccentricity of hyperbola (i), then $b^2=a^2\left(e^2-1\right)$
or $\frac{1}{e^2}=\frac{a^2}{\left(a^2+b^2\right)}$ ....(iii)
Similarly if $e^{\prime}$ is eccentricity of conjugate (ii), then $a^2=b^2\left(e^{\prime 2}-1\right)$ or $\frac{1}{e^{\prime 2}}=\frac{b^2}{\left(a^2+b^2\right)}$ .....(iv)
Adding (iii) and (iv),
\(\left(\frac{1}{e}\right)^2+\left(\frac{1}{e^1}\right)^2=1\)
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